What is the DFT of DFT of discrete signal
up vote
1
down vote
favorite
What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?
discrete-signals fourier-transform dft
add a comment |
up vote
1
down vote
favorite
What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?
discrete-signals fourier-transform dft
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?
discrete-signals fourier-transform dft
What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?
discrete-signals fourier-transform dft
discrete-signals fourier-transform dft
asked 8 hours ago
Mert Ege
133
133
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
let
$$begin{align}
X[k] &= mathcal{DFT} Big{ x[n] Big} \
&triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
end{align} $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$
then, if the DFT is defined the most common way (as above):
$$ Y[n] = N cdot x[-n] $$
where periodicity is implied: $x[n+N]=x[n]$ for all $n$.
add a comment |
up vote
1
down vote
Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).
Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
7 hours ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
let
$$begin{align}
X[k] &= mathcal{DFT} Big{ x[n] Big} \
&triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
end{align} $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$
then, if the DFT is defined the most common way (as above):
$$ Y[n] = N cdot x[-n] $$
where periodicity is implied: $x[n+N]=x[n]$ for all $n$.
add a comment |
up vote
4
down vote
let
$$begin{align}
X[k] &= mathcal{DFT} Big{ x[n] Big} \
&triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
end{align} $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$
then, if the DFT is defined the most common way (as above):
$$ Y[n] = N cdot x[-n] $$
where periodicity is implied: $x[n+N]=x[n]$ for all $n$.
add a comment |
up vote
4
down vote
up vote
4
down vote
let
$$begin{align}
X[k] &= mathcal{DFT} Big{ x[n] Big} \
&triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
end{align} $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$
then, if the DFT is defined the most common way (as above):
$$ Y[n] = N cdot x[-n] $$
where periodicity is implied: $x[n+N]=x[n]$ for all $n$.
let
$$begin{align}
X[k] &= mathcal{DFT} Big{ x[n] Big} \
&triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
end{align} $$
and
$$ y[n] triangleq X[n] $$
(note the substitution of $n$ in for $k$.) then
$$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$
then, if the DFT is defined the most common way (as above):
$$ Y[n] = N cdot x[-n] $$
where periodicity is implied: $x[n+N]=x[n]$ for all $n$.
answered 7 hours ago
robert bristow-johnson
10.4k31448
10.4k31448
add a comment |
add a comment |
up vote
1
down vote
Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).
Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
7 hours ago
add a comment |
up vote
1
down vote
Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).
Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
7 hours ago
add a comment |
up vote
1
down vote
up vote
1
down vote
Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).
Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).
Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).
Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).
edited 7 hours ago
answered 7 hours ago
hotpaw2
25.5k53472
25.5k53472
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
7 hours ago
add a comment |
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
7 hours ago
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
7 hours ago
shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
– robert bristow-johnson
7 hours ago
add a comment |
Thanks for contributing an answer to Signal Processing Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdsp.stackexchange.com%2fquestions%2f53959%2fwhat-is-the-dft-of-dft-of-discrete-signal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown